So while messing around with the Binet Formula for the Fibonacci numbers, I decided to input the Complex Numbers into it. And while doing so I noticed something. The Zero's of this sequence are all on a line (I'll link the graph later + Image.) So I was just curios on why the Zero's show up in a line?
https://samuelj.li/complex-function-plotter/#((phi%5Ez)-(-1%2Fphi)%5Ez)%2F(sqrt(5))
We want the zeros in the complex plotter. If one solves:
$$\frac{\phi^z-(-1/\phi)^z}{\sqrt{5}}=0\iff \phi^z=\left(-\frac1\phi\right)^z$$
by taking the complex logarithm:
$$z\ln(\phi)=z\ln\left(-\frac1\phi\right)+2\pi i n$$
Now factor and combine logarithms to get:
$$(\ln(\phi)-(\ln(-1)-\ln(\phi)))z=(2\ln(\phi)-\pi i)z=2\pi i n\implies z=\frac{2\pi in}{2\ln(\phi)-\pi i}$$
Plotting the points $(x,y)=(\operatorname{Re}(z),\operatorname{Im}(z))$ for the roots and using $\frac{\text{rise}}{\text{run}}$ slope formula gives that the roots lie on:
$$y=\frac{\operatorname{Im}\left(\frac{2\pi i}{2\ln(\phi)-\pi i}\right)}{\operatorname{Re}\left(\frac{2\pi i}{2\ln(\phi)-\pi i}\right)}x=-\frac2\pi\ln(\phi)x$$
The roots lie on a line because they are multiples of a complex number from which the slope can be found.